What is the derivative of y=4^{-x} ?

The derivative of y=4^{-x} is -ln(2)*2^{-2x+1}

What is the first derivative of y=4^{-x} ?

The first derivative of y=4^{-x} is -ln(2)*2^{-2x+1}

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derivative of

y = 4^{- x}

- ln (2) \cdot 2^{- 2 x + 1}

Solution steps

\frac{d}{d x} (4^{- x})

Apply exponent rule:

a^{b} = e^{b l n (a)}

4^{- x} = e^{(- x) l n (4)}

= \frac{d}{d x} (e^{(- x) l n (4)})

Apply the chain rule:

e^{(- x) l n (4)} \frac{d}{d x} ((- x) ln (4))

= e^{(- x) l n (4)} \frac{d}{d x} ((- x) ln (4))

\frac{d}{d x} ((- x) ln (4)) = - 2 ln (2)

= e^{(- x) l n (4)} (- 2 ln (2))

Simplify

e^{(- x) l n (4)} (- 2 ln (2)) : - ln (2) \cdot 2^{- 2 x + 1}

= - ln (2) \cdot 2^{- 2 x + 1}

\frac{d}{d y} (e^{5 y})

f (x) = sin (2 x), at x = \frac{π}{4}

h \to + 0 lim (\frac{( ( x + h ) - x )}{h})

\frac{d}{d x} ((\frac{a + b x ^{n}}{a - b x ^{n}})^{m})

x \to \infty lim (\frac{8 x x ^{4}}{9})

What is the derivative of y=4^{-x} ?
The derivative of y=4^{-x} is -ln(2)*2^{-2x+1}
What is the first derivative of y=4^{-x} ?
The first derivative of y=4^{-x} is -ln(2)*2^{-2x+1}